b. A cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0.02 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.05 cents per square centimeter. Find the dimensions for the can that will minimize production cost:
i) Radius of the can.
ii) Height of the can.
iii) Minimum cost.
Complete solution pls thanks!
This is just like your earlier problem. Where do you get stuck?
I dont know how to solve this to be honest... So yeah
To find the dimensions for the can that will minimize production cost, we need to minimize the cost as a function of the dimensions.
Let's denote the radius of the can as "r" and the height of the can as "h".
The total cost of the can is the sum of the cost of material for the sides and the cost of material for the top and bottom.
1) Cost of material for the sides:
The side of the can is in the shape of a rectangle and its area can be calculated as the height multiplied by the perimeter (circumference) of the circular base of the can, which is 2πr. Therefore, the area of the side is given by A_side = 2πrh.
The cost of the material for the sides is calculated as the area multiplied by the cost per square centimeter, which is 0.02 cents per square centimeter. Therefore, the cost of the material for the sides is C_side = 0.02 * A_side = 0.02 * 2πrh.
2) Cost of material for the top and bottom:
The top and bottom of the can are in the shape of circles, and their combined area can be calculated as 2 times the area of a single circle with radius r. Therefore, the area of the top and bottom is given by A_top_bottom = 2πr^2.
The cost of the material for the top and bottom is calculated as the area multiplied by the cost per square centimeter, which is 0.05 cents per square centimeter. Therefore, the cost of the material for the top and bottom is C_top_bottom = 0.05 * A_top_bottom = 0.05 * 2πr^2.
The total cost of the can is given by the sum of the cost of the material for the sides and the cost of the material for the top and bottom:
Cost = C_side + C_top_bottom = 0.02 * 2πrh + 0.05 * 2πr^2.
Now we need to find the dimensions that minimize this cost. We can do this by taking partial derivatives with respect to r and h, setting them equal to zero, and solving for r and h.
Taking the partial derivative with respect to r:
d(Cost)/dr = 0.04πh + 0.1πr = 0.
Taking the partial derivative with respect to h:
d(Cost)/dh = 0.02πr = 0.
Setting both derivatives to zero and solving for r and h, we have:
0.04πh + 0.1πr = 0,
0.02πr = 0.
From the second equation, we get r = 0, which means there is no cost associated with the top and bottom of the can for r = 0. However, a can with no radius (a flat sheet) is not practical. Therefore, we consider the first equation and solve for h:
0.04πh + 0.1πr = 0,
0.04h + 0.1r = 0,
0.04h = -0.1r,
h = -2.5r.
Substituting this value of h into the equation for the area of the side, we get:
A_side = 2πrh,
A_side = 2πr(-2.5r),
A_side = -5πr^2.
Now we substitute this value of A_side into the equation for the cost of the can:
Cost = 0.02 * A_side + 0.05 * 2πr^2,
Cost = 0.02 * (-5πr^2) + 0.05 * 2πr^2,
Cost = -0.1πr^2 + 0.1πr^2,
Cost = 0.
It turns out that the cost is zero for any value of r. This implies that the cost is minimized when the can has no sides (r = 0), which is not a practical solution.
Therefore, for this particular problem, there is no minimum cost solution that satisfies the constraints. The cost is zero for any value of r, which is not achievable with real dimensions.